Calculus: Early Transcendentals 8th Edition

by
Stewart, James

Answer

$y'=-\dfrac{1}{s^{2}}+\dfrac{3}{2\sqrt{s^{5}}}$

Work Step by Step

$y=\dfrac{s-\sqrt{s}}{s^{2}}$
Rewrite the function like this:
$y=s^{-2}(s-s^{1/2})=s^{-1}-s^{-3/2}$
Differentiate each term:
$y'=-s^{-2}-(-\dfrac{3}{2})s^{-5/2}=-s^{-2}+\dfrac{3}{2}s^{-5/2}=-\dfrac{1}{s^{2}}+\dfrac{3}{2s^{5/2}}=-\dfrac{1}{s^{2}}+\dfrac{3}{2\sqrt{s^{5}}}$
Please note that there was an easier way to differentiate the function than using the quotient rule